Tim and Gabriel started on a 55-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 45 km, Tim cycled faster than Gabriel. He arrived at the finishing point 24 minutes before Gabriel who was 10 km behind him. Gabriel did not change his speed throughout and completed it at 07 30.

1. At what time did the trip begin? Give the answer in 12-hour format.
2. What was Tim's average speed for the remaining 45 km of the trip in km/h?

(a)
60 min = 1 h
24 min =²⁴₆₀ h = ²₅ 

Gabriel's average speed
= 10 ÷ ²₅
= 25 km/h

Time that Gabriel took for the trip
= 55 ÷ 25
= 2⁵₂₅ h
= 2¹₅ h
= 2 h 12 min

2 h 12 min before 07 30 is 05 18.

(b)
To find the time that Tim took for the remaining 45 km of the trip, we need to use the time that Gabriel took subtract that of the time taken for first 10 km and the time that Tim was faster than Gabriel.

Time that Tim took for the remaining 45 km of the trip
= 2¹₅ - ²₅ - ²₅
= 1²₅ h

Tim's average speed
= 45 ÷ 1²₅
= 32¹₇ km/h

Answer(s): (a) 05 18; (b) 32¹₇ km/h